Getting the variance of $X$ from $Var(\ln(X))$ I've a multiplicative model for which $Y = X*Z$, for which $Y$ and $Z$ is known. I want this model to be additive (therefore using logarithms) to figure out the variance of $X$. I have $$ln(Y) = ln(X) + ln(Z)$$ and so $$VAR(ln(Y)) = VAR(ln(X)) + VAR(ln(Z))$$
and finally $$VAR(ln(X) = VAR(ln(Y)) - VAR(ln(Z))$$.
I'm not looking for a way to calculate $X$, only the variance of $X$. I now know $VAR(ln(X))$, but is there I can "revert" this variance to get the variance for $X$?
 A: This is not possible in general.
This is because it's possible to find different random variables $X$ and $Y$ where $V(\log(X)) = V(\log(Y))$ but $V(X) \ne V(Y)$, so even if you know $V(\log(X))$ you cannot recover the original variance $V(X)$ (at least not without making some additional assumptions).
For example, suppose we know that $X \sim \text{Lognormal} (\mu, \sigma^2)$, and hence $\log (X) \sim \text{Normal}(\mu, \sigma^2)$. The variance of a log-normal random variable is
$$V(X) = (\exp(\sigma^2) - 1)\exp(2\mu + \sigma^2).$$
This means that even if we know $V(\log(X)) = \sigma^2 = \log(2) \approx 0.693$, for example, the variance of $X$ still depends on $\mu$ through the equation
\begin{align}
V(X) & = (\exp(\log(2)) - 1)\exp(2\mu + \log(2)) \\
& = (2 - 1)\exp(2\mu) \exp(\log(2)) \\
& = 2 \exp(2\mu).
\end{align}
Hence, the variance of $X$ could be literally any positive number, even though we know $V(\log(X)) = \log(2)$. For example, it's possible that $\mu = 0$ which would give $V(X) = 2$, but it's also possible that $\mu = 1$ which would give $V(X) = 2 \exp(2) \approx 14.778$, and just knowing $V(\log(X))$ is not enough to distinguish between these possibilities.
A: The answer by @ericperkerson is completely right, you cannot in general find $\text{Var}(X)$ using only $\text{Var}(\ln X)$, but I would like to bring out the delta method.  If $X\sim\mathcal N(\mu,\sigma^2)$ and $\sigma$ is small enough*, we have the following approximation:
$$f(X)\overset{\text{approx.}}\sim\mathcal N(f(\mu), f'(\mu)^2\cdot\sigma^2)$$
In the case of $\ln X$, this gives us:
$$\text{Var}(\ln X)\approx\frac{\text{Var}(X)}{\mathbb E(X)^2}$$
So, if besides the variance of $X$ you also know its expectation, you can find this simple approximation for $\text{Var}(\ln X)$. How good this approximation is likely to depend on the shape and standard deviation of your distribution.
* small enough here means $\sigma^2<<\frac{1}{f''(\mu)}$, so that our approximation can safely ignore any curvature of $f(x)$.
